What are the DRs of vectors parallel to (2, -1, 1) and (3, 4, -1)?

(1, 5, -2)

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(-2, -5, 2)

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(-1, 5, 2)

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(-1, -5, -2)

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Solution

The correct option is A
(1, 5, -2)

The correct option is A.
Explanation of the correct option: -
$L e t a, b a n d c b e t h e d i r e c t i o n v e c t o r s o f t h e v e c t o r t h a t i s p e r p e n d i c u l a r t o t h e v e c t o r s h a v i n g d i r e c t i o n r a t i o s a s (2, - 1, 1) & (3, 4, - 1) . S o, 2 a - b + c = 0 & 3 a + 4 b - c = 0 = (3 - 2), (4 + 1), (- 1 - 1) = (1, 5, - 2) = (a, b, c) T h e r e f o r e, (a, b, c) = (1, 5, - 2)$
Explanation of the incorrect options: -
Option B: -
$L e t a, b a n d c b e t h e d i r e c t i o n v e c t o r s o f t h e v e c t o r t h a t i s p e r p e n d i c u l a r t o t h e v e c t o r s h a v i n g d i r e c t i o n r a t i o s a s (2, - 1, 1) & (3, 4, - 1) . S o, 2 a - b + c = 0 & 3 a + 4 b - c = 0 = (3 - 2), (4 + 1), (- 1 - 1) = (1, 5, - 2) = (a, b, c) T h e r e f o r e, (a, b, c) = (1, 5, - 2)$
Option C: -
$L e t a, b a n d c b e t h e d i r e c t i o n v e c t o r s o f t h e v e c t o r t h a t i s p e r p e n d i c u l a r t o t h e v e c t o r s h a v i n g d i r e c t i o n r a t i o s a s (2, - 1, 1) & (3, 4, - 1) . S o, 2 a - b + c = 0 & 3 a + 4 b - c = 0 = (3 - 2), (4 + 1), (- 1 - 1) = (1, 5, - 2) = (a, b, c) T h e r e f o r e, (a, b, c) = (1, 5, - 2)$
Option D: -
$L e t a, b a n d c b e t h e d i r e c t i o n v e c t o r s o f t h e v e c t o r t h a t i s p e r p e n d i c u l a r t o t h e v e c t o r s h a v i n g d i r e c t i o n r a t i o s a s (2, - 1, 1) & (3, 4, - 1) . S o, 2 a - b + c = 0 & 3 a + 4 b - c = 0 = (3 - 2), (4 + 1), (- 1 - 1) = (1, 5, - 2) = (a, b, c) T h e r e f o r e, (a, b, c) = (1, 5, - 2)$
Hence, the DRs vectors parallel to the (2,-1,1) and (3,4,-1) is (1,5,-2)

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